Whakaoti i te 2a^3-3a^2+1=0

$$2 a ^ { 3 } - 3 a ^ { 2 } + 1 = 0$$

Whakaoti mō a

$$a=1$$

$$a=-\frac{1}{2}=-0.5$$

Ngā Upane Otinga

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau 1, ā, ka wehea e q te whakarea arahanga 2. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{1}{2},±1$$

Kimihia tētahi pūtake pērā mā te whakamātau i ngā uara tau tōpū katoa, e tīmata ana i te mea iti rawa mā te uara pū. Mēnā kāore he pūtake tau tōpū e kitea, whakamātauria ngā hautanga.

$$a=1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te a-k o te pūrau mō ia pūtake k. Whakawehea te 2a^{3}-3a^{2}+1 ki te a-1, kia riro ko 2a^{2}-a-1. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$2a^{2}-a-1=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 2 mō te a, te -1 mō te b, me te -1 mō te c i te ture pūrua.

$$a=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Mahia ngā tātaitai.

$$a=\frac{1±3}{4}$$

Whakaotia te whārite 2a^{2}-a-1=0 ina he tōrunga te ±, ina he tōraro te ±.

$$a=-\frac{1}{2} a=1$$

Rārangitia ngā otinga katoa i kitea.

$$a=1 a=-\frac{1}{2}$$